Problem: The lifespans of gorillas in a particular zoo are normally distributed. The average gorilla lives $23.2$ years; the standard deviation is $4.9$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a gorilla living less than $28.1$ years.
Solution: $23.2$ $18.3$ $28.1$ $13.4$ $33$ $8.5$ $37.9$ $68\%$ $16\%$ $16\%$ We know the lifespans are normally distributed with an average lifespan of $23.2$ years. We know the standard deviation is $4.9$ years, so one standard deviation below the mean is $18.3$ years and one standard deviation above the mean is $28.1$ years. Two standard deviations below the mean is $13.4$ years and two standard deviations above the mean is $33$ years. Three standard deviations below the mean is $8.5$ years and three standard deviations above the mean is $37.9$ years. We are interested in the probability of a gorilla living less than $28.1$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $68\%$ of the gorillas will have lifespans within 1 standard deviation of the average lifespan. The remaining $32\%$ of the gorillas will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({16\%})$ will live less than $18.3$ years and the other half $({16\%})$ will live longer than $28.1$ years. The probability of a particular gorilla living less than $28.1$ years is ${68\%} + {16\%}$, or $84\%$.